Integrand size = 24, antiderivative size = 17 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=-13-e^x+2 x-\log (x)+\log (\log (x)) \]
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Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6874, 2225, 45, 2339, 29} \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 x-e^x-\log (x)+\log (\log (x)) \]
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Rule 29
Rule 45
Rule 2225
Rule 2339
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-e^x+\frac {1-\log (x)+2 x \log (x)}{x \log (x)}\right ) \, dx \\ & = -\int e^x \, dx+\int \frac {1-\log (x)+2 x \log (x)}{x \log (x)} \, dx \\ & = -e^x+\int \left (\frac {-1+2 x}{x}+\frac {1}{x \log (x)}\right ) \, dx \\ & = -e^x+\int \frac {-1+2 x}{x} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -e^x+\int \left (2-\frac {1}{x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -e^x+2 x-\log (x)+\log (\log (x)) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=-e^x+2 x-\log (x)+\log (\log (x)) \]
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Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94
method | result | size |
default | \(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\) | \(16\) |
norman | \(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\) | \(16\) |
risch | \(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\) | \(16\) |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\) | \(16\) |
parts | \(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\) | \(16\) |
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none
Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 x - e^{x} - \log {\left (x \right )} + \log {\left (\log {\left (x \right )} \right )} \]
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none
Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 15.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2\,x+\ln \left (\ln \left (x\right )\right )-{\mathrm {e}}^x-\ln \left (x\right ) \]
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